Question

The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.3 15.3 15.7 15.7 15.3 15.9 15.3 15.9 Construct a 98% confidence interval for the mean amount of juice in all such bottles

Answer 1

Answer:

$15.55-2.997\frac{0.278}{\sqrt{8}}=15.26$    

$15.55+2.997\frac{0.278}{\sqrt{8}}=15.84$    

The 98% confidence interval would be given by (15.26;15.84)    

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

We can calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=15.55$

The sample deviation calculated $s=0.278$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=8-1=7$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,7)".And we see that $t_{\alpha/2}=2.997$

And the confidence interval is given by:

$15.55-2.997\frac{0.278}{\sqrt{8}}=15.26$    

$15.55+2.997\frac{0.278}{\sqrt{8}}=15.84$    

The 98% confidence interval would be given by (15.26;15.84)